# How do I derive the formula for $\sum\limits_{x = 3}^{\infty} 1.536 (x^2) \left(\frac 5 8\right)^x$?

How would I find the summation of

$$\sum\limits_{x = 3}^{\infty} 1.536 (x^2) \left(\frac 5 8\right)^x$$

Would I have to take the 2nd derivative of $(1/1-x)$?

Yes, but it’s helpful to insert an intermediate step. Starting with

$$\frac1{1-x}=f(x)=\sum_{n\ge 0}x^n\;,$$

you get

$$\frac1{(1-x)^2}=f\,'(x)=\sum_{n\ge 0}nx^{n-1}\;,$$

and therefore

$$\frac{x}{(1-x)^2}=xf\,'(x)=x\sum_{n\ge 0}nx^{n-1}=\sum_{n\ge 0}nx^n\;.$$

Now what happens if you differentiate a second time and then multiply by $x$ again?

• Would I get -2x^2/(1-x)^3? – Jesus Oct 14 '13 at 2:21
• @Jesus: Your differentiation is a bit off. I get $$\frac{1-x}{(1-x)^3}$$ for the derivative of $$\frac{x}{(1-x)^2}\;.$$ – Brian M. Scott Oct 14 '13 at 2:24
• Wouldn't the answer be negative? Otherwise, x is the ratio (5/8) and I would need to subtract the first 2 terms from it which would be (1.536 * 5/8) and (1.536 * 4 * (5/8)^2)? – Jesus Oct 14 '13 at 2:30
• @Jesus: No, the derivative is not negative. Yes, you’ll substitute $\frac58$ for $x$, and you’ll have to subtract the first three terms, the ones for $n=0$, $n=1$, and $n=2$. – Brian M. Scott Oct 14 '13 at 2:33
• WolframAlpha gives me a negative derivative wolframalpha.com/input/?i=derive+x%2F%281-x%29%5E2... when n=0, wouldn't just be 0? Are the other 2 terms correct for when n=1 and n=2? – Jesus Oct 14 '13 at 2:36